我试图为我开发的AI递归构建二叉树。 我尝试构建一棵树，但一切都回来了。语言是Java，我使用的是Eclipse。另外，如果这意味着什么，我会在Mac上。树应作为二叉树返回，其中节点已实例化但没有任何内容。
public class DecisionTree {
//build a generic, empty, tree
//building binary
Root r = new Root();
public void build() //ok
{
Node lhs = new Node();
Node rhs = new Node();
lhs = new Node();
rhs = new Node();
r.lhs = lhs;
r.rhs = rhs;
lhs.parent = r;
rhs.parent = r;
builtRecursion(lhs, 1);
builtRecursion(rhs, 1);
outputTree();
int ctr = 1; //levels of tree
}
public int builtRecursion(Node n, int ctr)
{
Node lhs = new Node();
Node rhs = new Node();
ctr++;
System.out.println("built recursion ctr is " + ctr);
if (ctr > 10)
{
//leaf node
Behaviors behavior = new Behaviors();
Node node = behavior;
n.b = behavior;
return 0;
}
n.lhs = lhs;
n.rhs = rhs;
lhs.parent = n;
rhs.parent = n;
builtRecursion(lhs, ctr);
builtRecursion(rhs, ctr);
return ctr;
}
public void outputTree()
{
if (r != null)
{
System.out.println("Root");
}
outputTreeRecursive(r);
}
public void outputTreeRecursive(Node n)
{
if (n.lhs != null)
{
System.out.println("------------------");
System.out.println("LHS");
outputTreeRecursive(n.lhs);
}
else { System.out.println("LHS is null");}
if (n.rhs != null)
{
System.out.println("-----------------");
System.out.println("RHS");
outputTreeRecursive(n.rhs);
}
else { System.out.println("RHS is null");}
System.out.println("-----------------");
}
}
ROOT CLASSS
package FLINCH;
public class Root extends Node {
Node lhs = new Node();
Node rhs = new Node();
}
NODE CLASS
package FLINCH;
import java.util.ArrayList;
import java.util.LinkedList;
public class Node {
Node lhs = null;
Node rhs = null;
Node parent = null;
Decider d = new Decider(this);
Behaviors b = null;
public LinkedList getSuccessors()
{
LinkedList list = new LinkedList();
list.add(lhs);
list.add(rhs);
return list;
}
}
输出
GetAction Running
Iterating through open list
Size of open list is 1
Peeked openLIst size is 1
Peeking throguh open list
Popping Open List
LHS is null
RHS is null
Number of children is 2
Children equals 2
Decider childrens loop
Child node is null
Iterating through children
Exception in thread "main" java.lang.NullPointerException
at FLINCH.A_Star_Search.search3(A_Star_Search.java:81)
at FLINCH.Soldier.search_behavior(Soldier.java:28)
at FLINCH.Main.getAction(Main.java:54)
at tests.GameVisualSimulationTest.main(GameVisualSimulationTest.java:52)
我希望这会有所帮助......
答案 0 :(得分：0)
我有一段代码可以用于BinaryTree
public class BinarySearchTree {
public static Node root;
public BinarySearchTree(){
this.root = null;
}
public void insert(int id){
Node newNode = new Node(id);
if(root==null){
root = newNode;
return;
}
Node current = root;
Node parent = null;
while(true){
parent = current;
if(id < current.data){
current = current.left;
if(current==null){
parent.left = newNode;
return;
}
}else{
current = current.right;
if(current==null){
parent.right = newNode;
return;
}
}
}
}
public boolean find(int id){
Node current = root;
while(current!=null){
if(current.data==id){
return true;
}else if(current.data > id){
current = current.left;
}else{
current = current.right;
}
}
return false;
}
public boolean delete(int id){
Node parent = root;
Node current = root;
boolean isLeftChild = false;
while(current.data!=id){
parent = current;
if(current.data > id){
isLeftChild = true;
current = current.left;
}else{
isLeftChild = false;
current = current.right;
}
if(current ==null){
return false;
}
}
//if i am here that means we have found the node
//Case 1: if node to be deleted has no children
if(current.left==null && current.right==null){
if(current==root){
root = null;
}
if(isLeftChild ==true){
parent.left = null;
}else{
parent.right = null;
}
}
//Case 2 : if node to be deleted has only one child
else if(current.right==null){
if(current==root){
root = current.left;
}else if(isLeftChild){
parent.left = current.left;
}else{
parent.right = current.left;
}
}
else if(current.left==null){
if(current==root){
root = current.right;
}else if(isLeftChild){
parent.left = current.right;
}else{
parent.right = current.right;
}
}else if(current.left!=null && current.right!=null){
//now we have found the minimum element in the right sub tree
Node successor = getSuccessor(current);
if(current==root){
root = successor;
}else if(isLeftChild){
parent.left = successor;
}else{
parent.right = successor;
}
successor.left = current.left;
}
return true;
}
public Node getSuccessor(Node deleleNode){
Node successsor =null;
Node successsorParent =null;
Node current = deleleNode.right;
while(current!=null){
successsorParent = successsor;
successsor = current;
current = current.left;
}
//check if successor has the right child, it cannot have left child for sure
// if it does have the right child, add it to the left of successorParent.
// successsorParent
if(successsor!=deleleNode.right){
successsorParent.left = successsor.right;
successsor.right = deleleNode.right;
}
return successsor;
}
public void display(Node root){
if(root!=null){
display(root.left);
System.out.print(" " + root.data);
display(root.right);
}
}
public static void printInOrder(Node root){
if(root == null){
return;
}
printInOrder(root.left);
System.out.print(root.data+" ");
printInOrder(root.right);
}
public static void printPreOrder(Node root){
if(root == null){
return;
}
System.out.print(root.data+" ");
printPreOrder(root.left);
printPreOrder(root.right);
}
public static void printPostOrder(Node root){
if(root == null){
return;
}
printPostOrder(root.left);
printPostOrder(root.right);
System.out.print(root.data+" ");
}
public static void main(String arg[]){
BinarySearchTree b = new BinarySearchTree();
b.insert(3);b.insert(8);
b.insert(1);b.insert(4);b.insert(6);b.insert(2);b.insert(10);b.insert(9);
b.insert(20);b.insert(25);b.insert(15);b.insert(16);
System.out.println("Original Tree : ");
b.display(b.root);
System.out.println("");
System.out.println("Check whether Node with value 4 exists : " + b.find(4));
System.out.println("Delete Node with no children (2) : " + b.delete(2));
b.display(root);
System.out.println("\n Delete Node with one child (4) : " + b.delete(4));
b.display(root);
System.out.println("\n Delete Node with Two children (10) : " + b.delete(10));
b.display(root);
System.out.println();
System.out.println("********* Printing In Order *********");
printInOrder(root);
System.out.println();
System.out.println("********* Printing Pre Order *********");
printPreOrder(root);
System.out.println();
System.out.println("********* Printing Post Order *********");
printPostOrder(root);
}
}
class Node{
int data;
Node left;
Node right;
public Node(int data){
this.data = data;
left = null;
right = null;
}
}
答案 1 :(得分：0)
当您使用buildRecursion调用ctr = 1时，您可能意味着您想要构建一个只有一个额外级别的树，并且根据您希望构建叶节点的注释，条件需要被修改。在你的情况下，条件应该是：
if (ctr == 1)
我对您的功能进行了一些更改以获得更好的输出：
public void builtRecursion(Node n, int ctr)
{
System.out.println("built recursion ctr is " + ctr);
if (ctr == 1)
{
//leaf node
Behaviors behavior = new Behaviors();
n.b = behavior;
return;
}
Node lhs = new Node();
Node rhs = new Node();
n.lhs = lhs;
n.rhs = rhs;
lhs.parent = n;
rhs.parent = n;
builtRecursion(lhs, ctr--);
builtRecursion(rhs, ctr--);
}
关于的问题我尝试构建一个树，但一切都回来了，在你的outputTree和outputRecursion中你不打印任何东西而不是"-----"，"LHS"，"RHS"如果您到达没有左或右节点的地方，您将打印＆＃34; LHS / RHS为空＆＃34;但是您应该知道，对于叶节点，这是一种可接受的行为，因此当您到达叶节点时，您应该打印它们的值。但是，您可以将outputTreeRecursive更改为以下内容：
public void outputTreeRecursive(Node n)
{
if (n.lhs != null)
{
System.out.println("------------------");
System.out.println("LHS");
outputTreeRecursive(n.lhs);
}
else {
System.out.println("LHS is null");
if(n.b != null)
System.out.println("Leaf node");
}
if (n.rhs != null)
{
System.out.println("-----------------");
System.out.println("RHS");
outputTreeRecursive(n.rhs);
}
else {
System.out.println("RHS is null");
if(n.b != null)
System.out.println("Leaf node");
}
System.out.println("-----------------");
}
现在可能你对树有更好的了解